\(n_O=\dfrac{4,8}{16}=0,3\left(mol\right)\)
Ta có: \(n_{Fe\left(OH\right)_3}=\dfrac{1}{3}n_O=\dfrac{1}{3}\times0,3=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe\left(OH\right)_3}=0,1\times107=10,7\left(g\right)\)
Ta có: \(n_H=3n_{Fe\left(OH\right)_3}=3\times0,1=0,3\left(mol\right)\)
\(\Rightarrow m_H=0,3\times1=0,3\left(g\right)\)